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for $a_n = \frac {n!!}{(n+1)!!}$, prove that $\lim a_n = 0$
putting $a_n$ into logarithm, we get $$\ln a_n=\sum_{k=1}^n \ln \frac k {k+1} = -\sum_{k=1}^n \ln (1+\frac1 k)$$

by taylor series,

$$\ln(1+1)=\frac 1 1 - \frac 1 2 + \frac 1 3+...$$ $$\ln(1+\frac1 2)=\frac 1 2 - \frac 1 8 + \frac 1 {24}-...$$ $$...$$ $$\ln(1+\frac1 k)=\frac 1 k - \frac 1 {2k^2} + \frac 1 {3k^3}-...$$

if "$\lim \ln a_n$" exists, $\sum_1^\infty \ln(1+\frac 1 k)$ (absolutely) converges. so we know that any rearrangement series of $\ln(1+\frac 1 k)$ converges and has the same value.

furthermore, $\ln(1+\frac 1 k) =\sum_1^\infty\frac {(-1)^{k+1}} k$ converges absolutely if $k>1$

so, by assuning $\ln a_n$ converges, expand $\ln(1+\frac 1 k)$ by taylor series for $k>1$ and rearrange column by column. $$\sum_{k=1}^{\infty} \ln (1+\frac1 k)=(\ln2 + (\frac1 2 + \frac 1 3 + ...) - \frac 1 2 (\frac 1 4 + \frac 1 9 + ...) +\frac1 3(\frac 1 8 +\frac 1 {27} +...)+...)$$ however the second term $(\frac 1 2 + \frac 1 3 +...)$ doesn't converge while all the other terms converge. this contradiction proves that $\ln a_n$ doesn't converge. so we have

$$\lim_{n\to\infty}\ln a_n=-\sum_{k=1}^{\infty} \ln (1+\frac1 k)=-\infty$$

which implies $\lim a_n =0$.

  1. Is this proof correct?
  2. Can you give me more simple proof?
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Your formula for $a_n$ is wrong, because $m!! = \prod_{k=0}^{\lfloor (m-1)/2 \rfloor} (m-2k) $. Therefore $a_n =\dfrac{n!!}{(n+1)!!} =\dfrac{\prod_{k=0}^{\lfloor (n-1)/2 \rfloor} (n-2k)}{\prod_{k=0}^{\lfloor /2 \rfloor} (n+1-2k)} $.

I'll show for even $n$ and leave odd $n$ up to you.

$\begin{array}\\ a_{2n} &=\dfrac{(2n)!!}{(2n+1)!!}\\ &=\dfrac{\prod_{k=0}^{n-1} (2n-2k)}{\prod_{k=0}^{n} (2n+1-2k)}\\ &=\prod_{k=0}^{n-1} \left(\dfrac{2n-2k}{2n+1-2k}\right)\\ &=\prod_{k=0}^{n-1} \left(1-\dfrac{1}{2n+1-2k}\right)\\ &=\prod_{k=1}^{n} \left(1-\dfrac{1}{2k+1}\right) \qquad\text{ putting } n-k \text{ for } k\\ \end{array} $

Since $\sum \dfrac{1}{2k+1}$ diverges, the product goes to zero.

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Note that for the odd terms, we have

$$a_{2n-1}=\frac{(2n-1)!!}{(2n)!!}=\frac{(2n)!}{4^n\,(n!)^2}$$

Using Stirling's Formula, we have

$$\begin{align} a_{2n-1}\sim \frac{\sqrt{4\pi n}\left(\frac{2n}{e}\right)^{2n}}{4^n(2\pi n)\left(\frac{n}{e}\right)^{2n}}=\sqrt{\frac{\pi}{n}} \end{align}$$

Therefore, $\lim_{n\to \infty}a_{2n-1}=0$.


We also have for the even terms

$$\begin{align} a_{2n}&=\frac{(2n)!!}{(2n+1)!!}\\\\ &=\frac{1}{(2n+1)a_{2n-1}}\\\\ &\sim\frac{\sqrt{n}}{\sqrt{\pi}(2n+1)}\to 0 \end{align}$$

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  • $\begingroup$ @MrTanorus Please let me know how to improve my answer. I really want to give you the best answer I can. -Mark $\endgroup$
    – Mark Viola
    Mar 25 '17 at 22:21

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